Is this question on non singular matrix valid? > If A is a non singular matrix satisfying $AB- BA = A$, then prove that $\det(B+I) =\det(B-I)$ But is this question even valid because if we take the determinant of both sides of the first equation we get: $\det(AB- BA)= \det(A) \implies \det(AB)- \det(BA)= \det A \implies \det (A)= 0 $ but $\det (A) \ne 0$ for A to be non singular. So I think this question is wrong. Can someone please verify?

The determinant is multiplicative, not additive.

In this case, we must bring up multiplicativity somewhere, clearly. Note that $AB - BA = A$. To bring multiplicativity here, note that $AB = BA + A $, so that $\det(AB) = \det((B+I)A)$. Then, by multiplicativity, $\det A\det B = \det A \det(B+I)$.

Similarly, we have $AB -BA = A$ so $AB - A = BA$, so $\det (A(B-I)) = \det A \det B$. Consequently, we get that $\det A \det (B+I) = \det A \det(B-I)$. Why can we conclude that $\det A \
eq 0$? (Look at hypotheses). Now conclude.